What is the center ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 ?

The center ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 is (3,1)

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center ((x-3}{16})^2-\frac{(y-1)^2)/9 =1

center

(\frac{x - 3}{16})^{2} - \frac{( y - 1 ) ^{2}}{9} = 1

(3, 1)

Solution steps

(\frac{x - 3}{16})^{2} - \frac{( y - 1 ) ^{2}}{9} = 1

Rewrite

(\frac{x - 3}{16})^{2} - \frac{( y - 1 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{1 6 ^{2}} - \frac{( y - 1 ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, 1), a = 16, b = 3

And the center is:

(3, 1)

(x - y - 3) \cdot (x + y + 3) = 16

9 = 2 y - y^{2} - 6 \cdot (- x^{2})

4 x^{2} + 9 y^{2} + 3 x - 18 y^{2} + 37 = 0

9 y^{2} - x^{2} - 36 y - 6 x + 18 = 0

\frac{( x - 2 ) ^{2}}{9} - \frac{( y + 2 ) ^{2}}{9} = 1

What is the center ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 ?
The center ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 is (3,1)